Problem: $g(x)=\begin{cases} \dfrac1x&\text{for }x<-1 \\\\ 1+2x&\text{for }-1\leq x\leq0 \end{cases}$ Find $\lim_{x\to -1}g(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $1$ (Choice C) C $3$ (Choice D) D The limit doesn't exist.
Explanation: $x=-1$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to -1}g(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $-1$ from the left. We will use the fact that $g(x)=\dfrac1x$ for $x$ -values smaller than $-1$. $\begin{aligned} &\phantom{=}\lim_{x\to -1^-}g(x) \\\\ &=\lim_{x\to -1^-}\dfrac1x \\\\ &=\dfrac{1}{-1}&\gray{\text{Direct substitution}} \\\\ &=-1 \end{aligned}$ Let's find the limit as $x$ approaches $-1$ from the right. We will use the fact that $g(x)=1+2x$ for $x$ -values greater than $-1$. $\begin{aligned} &\phantom{=}\lim_{x\to -1^+}g(x) \\\\ &=\lim_{x\to -1^+}1+2x \\\\ &=1+2(-1)&\gray{\text{Direct substitution}} \\\\ &=-1 \end{aligned}$ The one-sided limits are both equal to $-1$. This means that $\lim_{x\to -1}g(x)=-1$.